题目描述 利用重载运算符，实现矩阵的加法，减法。 请重载加减运算符并自定义矩阵类完成相应的操作 输入 第一行为测试数据数 对于每组测试数据 第一行为矩阵的行数和列数 接下来分别为两个矩阵的各个元素 输出 输出矩阵的和与差

以下是一个可能的实现，其中矩阵类使用动态分配的二维数组来存储矩阵元素：

#include <iostream>
using namespace std;

class Matrix {
private:
    int rows;
    int cols;
    int **data;
public:
    Matrix(int r, int c) {
        rows = r;
        cols = c;
        data = new int*[rows];
        for (int i = 0; i < rows; i++) {
            data[i] = new int[cols];
        }
    }

    Matrix(const Matrix& other) {
        rows = other.rows;
        cols = other.cols;
        data = new int*[rows];
        for (int i = 0; i < rows; i++) {
            data[i] = new int[cols];
            for (int j = 0; j < cols; j++) {
                data[i][j] = other.data[i][j];
            }
        }
    }

    ~Matrix() {
        for (int i = 0; i < rows; i++) {
            delete[] data[i];
        }
        delete[] data;
    }

    Matrix operator+(const Matrix& other) const {
        if (rows != other.rows || cols != other.cols) {
            throw "Matrix dimensions do not match!";
        }
        Matrix result(rows, cols);
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                result.data[i][j] = data[i][j] + other.data[i][j];
            }
        }
        return result;
    }

    Matrix operator-(const Matrix& other) const {
        if (rows != other.rows || cols != other.cols) {
            throw "Matrix dimensions do not match!";
        }
        Matrix result(rows, cols);
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                result.data[i][j] = data[i][j] - other.data[i][j];
            }
        }
        return result;
    }

    friend ostream& operator<<(ostream& os, const Matrix& mat) {
        for (int i = 0; i < mat.rows; i++) {
            for (int j = 0; j < mat.cols; j++) {
                os << mat.data[i][j] << " ";
            }
            os << endl;
        }
        return os;
    }
};

int main() {
    int t;
    cin >> t;
    while (t--) {
        int m, n;
        cin >> m >> n;
        Matrix A(m, n);
        Matrix B(m, n);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                cin >> A.data[i][j];
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                cin >> B.data[i][j];
            }
        }
        try {
            cout << A + B << endl;
            cout << A - B << endl;
        } catch (const char* msg) {
            cerr << msg << endl;
        }
    }
    return 0;
}

在主函数中，首先读入测试数据的数量。对于每组测试数据，读入矩阵的行数和列数，然后分别读入两个矩阵的元素。然后使用重载的加减运算符求出矩阵的和与差，并输出结果。如果矩阵的维度不匹配，则抛出异常。注意，这里使用了 `try` 和 `catch` 来捕获异常，以便程序能够正常结束。